MATERIAL HANDLING— TRANSPORTATION OF SOLIDS 101 A similar equation for ascertaining the relative depreciation (annual) — Formula (22) — utilizes other values for constant K and a percentage of the factor D RDbe = F (~g§T + ° - 15^ Formula (22) Where, K = Constant = 8.0, buckets spaced 12 in. = 6.8, buckets spaced 15 in. = 5.6, buckets spaced 18 in. In charging up the various expenses contracted in operating a bucket elevator, depreciation expenses, which are relatively heavy while the equipment is in use and comparatively light when not in use, need only be charged against operating hours, provided the equipment is in use the greater part of 2,500 hr. during the year. Fixed charges, consisting of the burden of interest on investment, taxes, insurance, etc., may be taken at 8.5 per cent of the initial cost per year. Labor expense of 5 cts. per inch width of bucket will cover the necessary periodic inspections of equipment as well as the attention required to prevent unwieldy lumps passing to the buckets, etc. On account of the dust which works into the shaft bearings and the chain links and pins, the expense for incidental supplies is high, averaging close to 2 cts. per horsepower consumed. To illustrate the method of arriving at the total cost of operating a bucket elevator, the following typical example may be taken. Example. — Required the net operating cost of a 100-ft. double-strand (two chain) elevator handling sand weighing 120 Ib. per cubic foot. Elevator to have malleable-iron buckets and combination chains. Power delivered at elevator drive commanding a value of 3 cts. per horsepower. Elevator to be used 1,500 hr. per year. Capacity 25 tons per hour. Advisable speed of elevator (Table 11) 175 ft. per minute. Capacity, 25 = l^.™*™. Formula (19) 10.7/Sf = Iw* Let S = 12, then Iw* = 128.4. Select 6 by 4-in. bucket. w 6 X 4 X 4 X 175 X 120 1Q ... , , W « - _____ ----- = 18.00 tons per hour. Select 8 by 5-in. buckets, 18-in. spacing. w 8 X 5 X 5 X 175 X 120 or . . . , ., W — ---- n"AAA~^TVo~ — " = 25 tons per hour, required capacity. y,uuu x io Horsepower = -15 -- = 3.75— say 5 hp. Formula (20) 1.78 - o +1-0 X 5.X 8) = 1,673.2 Formula (21) . = 1.78 (^^X^100-*!?? + 0.15 X 1.9 X 5 X 8) - 280.2 Formula (22) Take market price at 25 per cent above relative costs. Cost = 1.25 X 1,673.2 = $2,091.50, say $2,100.00— initial cost.